int x^{2} [ sqrt{2} sin ( frac{pi}{4} + x ) + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The integral expression is given as,$I = \int x^{2} [ \sqrt{2} \sin ( \frac{\pi}{4} + x ) + e^{x} ] dx$ $\Rightarrow I = \int x^{2} [ \sqrt{2} ( \

Vedclass · Accepted Answer

$( x^{2} + 2 x - 2 ) \sin x + ( - x^{2} + 2 x + 2 ) \cos x + ( x^{2} - 2 x + 2 ) e^{x} + C$

Vedclass · Answer

(A) The integral expression is given as,$I = \int x^{2} [ \sqrt{2} \sin ( \frac{\pi}{4} + x ) + e^{x} ] dx$ $\Rightarrow I = \int x^{2} [ \sqrt{2} ( \sin \frac{\pi}{4} \cos x + \sin x \cos \frac{\pi}{4} ) + e^{x} ] dx$ $\Rightarrow I = \int x^{2} [ \sqrt{2} ( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x ) + e^{x} ] dx$ Further simplifying we get,$\Rightarrow I = \int x^{2} ( \cos x + \sin x + e^{x} ) dx$ $\Rightarrow I = \int x^{2} ( \cos x + \sin x ) dx + \int x^{2} e^{x} dx$ Using integration by parts,$\int u v dx = u \int v dx - \int \{ \frac{d u}{d x} \cdot \int v dx \} dx$: $\Rightarrow I = x^{2} ( \sin x - \cos x ) - \int 2 x ( \sin x - \cos x ) dx + ( x^{2} - 2 x + 2 ) e^{x} + C$ $\Rightarrow I = x^{2} ( \sin x - \cos x ) - 2 [ x ( - \cos x - \sin x ) - \int ( - \cos x - \sin x ) dx ] + ( x^{2} - 2 x + 2 ) e^{x} + C$ $\Rightarrow I = x^{2} ( \sin x - \cos x ) + 2 x ( \cos x + \sin x ) - 2 ( \sin x - \cos x ) + ( x^{2} - 2 x + 2 ) e^{x} + C$ $\Rightarrow I = ( x^{2} + 2 x - 2 ) \sin x + ( - x^{2} + 2 x + 2 ) \cos x + ( x^{2} - 2 x + 2 ) e^{x} + C$

$\int x^{2} [ \sqrt{2} \sin ( \frac{\pi}{4} + x ) + e^{x} ] dx =$

Similar Questions

If $\int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} = (\tan x)^A + C(\tan x)^B + K$,where $K$ is a constant of integration,then the value of $5(A+B+C)$ is equal to

Let $x > 0$ be a fixed real number. Then,the integral $\int \limits_0^{\infty} e^{-t}|x-t| d t$ is equal to

$\int \frac{\tan^{-1} x - \cot^{-1} x}{\tan^{-1} x + \cot^{-1} x} \, dx$ is equal to:

$\int (1+\tan^2 x)(1+2x \tan x) dx =$

Let $F(x)$ be an indefinite integral of $\sin ^2 x$.
$STATEMENT -1$ : The function $F(x)$ satisfies $F(x+\pi)=F(x)$ for all real $x$. because
$STATEMENT -2$: $\sin ^2(x+\pi)=\sin ^2 x$ for all real $x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module